Question

At some temperature and under a pressure of $5$ atm, $PC{l}_5$ is $20\%$ dissociated. Calculate the pressure at which $PC{l}_5$ will be $30\%$ dissociated, temperature remaining the same.

• A. $1.685$ atm
• B. $0.192$ atm
• C. $2.106$ atm
• D. $1.415$ atm

Answer 1

The correct option is C $2.106$ atm

At 5 atm pressure and 20% dissociation.

$PC{l}_5$ $\to$ $PC{l}_3$ + $C{l}_2$

Equibm mole 80 20 20

Equilibrium $(\frac{80}{120}\times 5)$ $(\frac{20}{120}\times 5)$ $(\frac{20}{120}\times 5)$

partial pressures

Hence, $K_p=\frac{P_{PC{l}_3}{P}_{C{l}_2}}{P_{PC{l}_5}}=\frac{5}{24}$.

At 30% dissociation,

$PC{l}_5$ $\to$ $PC{l}_3$ + $C{l}_2$

Equibm mole 70 30 30

Equilibrium $(\frac{70}{130}\times P)$ $(\frac{30}{130}\times P)$ $(\frac{30}{130}\times P)$

partial pressures

$K_p=\frac{P_{PC{l}_3}{P}_{C{l}_2}}{P_{PC{l}_5}}=\frac{9}{91}\times P$

Thus $\frac{9}{91}\times P=\frac{5}{24}$ or $P=2.106atm$.